Integral of x*cos(x)*sin(x)*dx dx

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$.

   Then $\operatorname{du}{\left(x \right)} = 1$.

   To find $v{\left(x \right)}$:

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

         $$\int u\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u\right)\, du = - \int u\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u\, du = \frac{u^{2}}{2}$$

            So, the result is: $- \frac{u^{2}}{2}$

         Now substitute $u$ back in:

         $$- \frac{\cos^{2}{\left(x \right)}}{2}$$

      Method #2

      1. Let $u = \sin{\left(x \right)}$.

         Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

         $$\int u\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u\, du = \frac{u^{2}}{2}$$

         Now substitute $u$ back in:

         $$\frac{\sin^{2}{\left(x \right)}}{2}$$

   Now evaluate the sub-integral.

2. The integral of a constant times a function is the constant times the integral of the function:

   $$\int \left(- \frac{\cos^{2}{\left(x \right)}}{2}\right)\, dx = - \frac{\int \cos^{2}{\left(x \right)}\, dx}{2}$$

   1. Rewrite the integrand:

      $$\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$$

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$$

         1. Let $u = 2 x$.

            Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

            $$\int \frac{\cos{\left(u \right)}}{4}\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$$

               1. The integral of cosine is sine:

                  $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

               So, the result is: $\frac{\sin{\left(u \right)}}{2}$

            Now substitute $u$ back in:

            $$\frac{\sin{\left(2 x \right)}}{2}$$

         So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

      The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$

   So, the result is: $- \frac{x}{4} - \frac{\sin{\left(2 x \right)}}{8}$

3. Now simplify:

   $$- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}$$

4. Add the constant of integration:

   $$- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}+ \mathrm{constant}$$

The answer is:

$$- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}+ \mathrm{constant}$$